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How can prove this integral

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4

1

$begingroup$

I was reading a textbook which these two equations posed . The second one was the result of the first one .
How can we say that ?

If we know :
$$int_{0^+}^{+infty} frac{sin(x)}{x} = frac{pi}{2}$$

How can we prove :
$$int_{0^+}^{+infty} left(frac{sin(x)}{x}right)^2 = frac{pi}{2}$$

Thanks in advance

real-analysis calculus integration

share|cite|improve this question

edited 5 hours ago

Alan Muniz

2,2711829

asked 5 hours ago

RezaReza

233

New contributor

Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

4

1

$begingroup$

I was reading a textbook which these two equations posed . The second one was the result of the first one .
How can we say that ?

If we know :
$$int_{0^+}^{+infty} frac{sin(x)}{x} = frac{pi}{2}$$

How can we prove :
$$int_{0^+}^{+infty} left(frac{sin(x)}{x}right)^2 = frac{pi}{2}$$

Thanks in advance

real-analysis calculus integration

share|cite|improve this question

edited 5 hours ago

Alan Muniz

2,2711829

asked 5 hours ago

RezaReza

233

New contributor

Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I was reading a textbook which these two equations posed . The second one was the result of the first one .
How can we say that ?

If we know :
$$int_{0^+}^{+infty} frac{sin(x)}{x} = frac{pi}{2}$$

How can we prove :
$$int_{0^+}^{+infty} left(frac{sin(x)}{x}right)^2 = frac{pi}{2}$$

Thanks in advance

real-analysis calculus integration

share|cite|improve this question

edited 5 hours ago

Alan Muniz

2,2711829

asked 5 hours ago

RezaReza

233

New contributor

Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 5 hours ago

Alan Muniz

2,2711829

asked 5 hours ago

RezaReza

233

New contributor

Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 5 hours ago

Alan Muniz

2,2711829

asked 5 hours ago

RezaReza

233

New contributor

Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 5 hours ago

Alan Muniz

2,2711829

edited 5 hours ago

Alan Muniz

2,2711829

asked 5 hours ago

RezaReza

233

New contributor

Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 5 hours ago

RezaReza

233

2 Answers
2

active

oldest

votes

6

$begingroup$

Use $int_{0}^{infty} frac{sin(x)}{x} = frac{pi}{2}$ and $sin (2x)= 2 sin(x) cos(x)$ to get

$$ (*) quadint_{0}^{infty} frac{sin(x) cos (x)}{x} =frac{pi}{4}.$$

Then use integration by parts in $(*)$ to derive

$$int_{0}^{infty} frac{sin^2(x)}{x^2} = frac{pi}{2}.$$

share|cite|improve this answer

answered 5 hours ago

FredFred

47k1848

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I can't figure it out how you derive from (*) to answer
  $endgroup$
  – Reza
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Reza: What is an antiderivative of the function cos? Can you derive the function defined for $x> 0$ by $f(x)=frac{sin x}{x}$ (product of functions)?
  $endgroup$
  – FDP
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

$$I(a)=int_{-infty}^{+infty}dfrac{sin^2ax}{x^2}mathrm dx\ dfrac{mathrm dI}{mathrm da}=int_{-infty}^{+infty}partial_a dfrac{sin^2ax}{x^2}mathrm dx=2int_{0}^{infty}dfrac{sin 2ax}{x}mathrm dx=pi\ I(a)=pi a implies int_{0}^{infty}dfrac{sin^2x}{x^2}mathrm dx =dfrac{1}{2}I(1)=dfrac{pi}{2}$$

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

Paras KhoslaParas Khosla

1,384219

$endgroup$

add a comment | 

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6

$begingroup$

Use $int_{0}^{infty} frac{sin(x)}{x} = frac{pi}{2}$ and $sin (2x)= 2 sin(x) cos(x)$ to get

$$ (*) quadint_{0}^{infty} frac{sin(x) cos (x)}{x} =frac{pi}{4}.$$

Then use integration by parts in $(*)$ to derive

$$int_{0}^{infty} frac{sin^2(x)}{x^2} = frac{pi}{2}.$$

share|cite|improve this answer

answered 5 hours ago

FredFred

47k1848

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I can't figure it out how you derive from (*) to answer
  $endgroup$
  – Reza
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Reza: What is an antiderivative of the function cos? Can you derive the function defined for $x> 0$ by $f(x)=frac{sin x}{x}$ (product of functions)?
  $endgroup$
  – FDP
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

6

$begingroup$

Use $int_{0}^{infty} frac{sin(x)}{x} = frac{pi}{2}$ and $sin (2x)= 2 sin(x) cos(x)$ to get

$$ (*) quadint_{0}^{infty} frac{sin(x) cos (x)}{x} =frac{pi}{4}.$$

Then use integration by parts in $(*)$ to derive

$$int_{0}^{infty} frac{sin^2(x)}{x^2} = frac{pi}{2}.$$

share|cite|improve this answer

answered 5 hours ago

FredFred

47k1848

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I can't figure it out how you derive from (*) to answer
  $endgroup$
  – Reza
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Reza: What is an antiderivative of the function cos? Can you derive the function defined for $x> 0$ by $f(x)=frac{sin x}{x}$ (product of functions)?
  $endgroup$
  – FDP
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Use $int_{0}^{infty} frac{sin(x)}{x} = frac{pi}{2}$ and $sin (2x)= 2 sin(x) cos(x)$ to get

$$ (*) quadint_{0}^{infty} frac{sin(x) cos (x)}{x} =frac{pi}{4}.$$

Then use integration by parts in $(*)$ to derive

$$int_{0}^{infty} frac{sin^2(x)}{x^2} = frac{pi}{2}.$$

share|cite|improve this answer

answered 5 hours ago

FredFred

47k1848

$endgroup$

share|cite|improve this answer

answered 5 hours ago

FredFred

47k1848

answered 5 hours ago

FredFred

47k1848

answered 5 hours ago

FredFred

47k1848

2

$begingroup$

$$I(a)=int_{-infty}^{+infty}dfrac{sin^2ax}{x^2}mathrm dx\ dfrac{mathrm dI}{mathrm da}=int_{-infty}^{+infty}partial_a dfrac{sin^2ax}{x^2}mathrm dx=2int_{0}^{infty}dfrac{sin 2ax}{x}mathrm dx=pi\ I(a)=pi a implies int_{0}^{infty}dfrac{sin^2x}{x^2}mathrm dx =dfrac{1}{2}I(1)=dfrac{pi}{2}$$

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

Paras KhoslaParas Khosla

1,384219

$endgroup$

add a comment | 

2

$begingroup$

$$I(a)=int_{-infty}^{+infty}dfrac{sin^2ax}{x^2}mathrm dx\ dfrac{mathrm dI}{mathrm da}=int_{-infty}^{+infty}partial_a dfrac{sin^2ax}{x^2}mathrm dx=2int_{0}^{infty}dfrac{sin 2ax}{x}mathrm dx=pi\ I(a)=pi a implies int_{0}^{infty}dfrac{sin^2x}{x^2}mathrm dx =dfrac{1}{2}I(1)=dfrac{pi}{2}$$

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

Paras KhoslaParas Khosla

1,384219

$endgroup$

add a comment | 

$begingroup$

$$I(a)=int_{-infty}^{+infty}dfrac{sin^2ax}{x^2}mathrm dx\ dfrac{mathrm dI}{mathrm da}=int_{-infty}^{+infty}partial_a dfrac{sin^2ax}{x^2}mathrm dx=2int_{0}^{infty}dfrac{sin 2ax}{x}mathrm dx=pi\ I(a)=pi a implies int_{0}^{infty}dfrac{sin^2x}{x^2}mathrm dx =dfrac{1}{2}I(1)=dfrac{pi}{2}$$

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

Paras KhoslaParas Khosla

1,384219

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 5 hours ago

Paras KhoslaParas Khosla

1,384219

answered 5 hours ago

Paras KhoslaParas Khosla

1,384219

answered 5 hours ago

Paras KhoslaParas Khosla

1,384219